Answer
$z = \pm r\sqrt {\cos 2\theta } $
Work Step by Step
We have ${z^2} = {x^2} - {y^2}$. The relations between rectangular and cylindrical coordinates are given by
$x = r\cos \theta $, ${\ \ }$ $y = r\sin \theta $, ${\ \ }$ $z=z$.
Substituting $x$ and $y$ in ${z^2} = {x^2} - {y^2}$ gives
${z^2} = {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta $
${z^2} = {r^2}\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)$
But $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $. So,
${z^2} = {r^2}\cos 2\theta $
$z = \pm r\sqrt {\cos 2\theta } $.
